Question

3. Let a: R+R be the regular curve defined by a(t) = (213, 121, 61). Find B, K and .

Answer 1

2: R3 R3 24 )= (2 t? 12t, 6+²) Tit)= x 2 (4) unit tangent vector Ñita I' (t) Unit normal vects 1 7' ll F (*) xÑ( do Bilarial Bingnala Brt)= Vects curvature keta 17') Iloco of kit) = lla't) x2"(t)! 110 Gill 2'Ct). (Q"(t) x 2" (4) 7= Tornion = la!(+)X2"C)/2 Now, 2(t) = (2+², 12t, 6t²) a' (t) = (6+², 12, 12t) 2"(+) = ( 12 t, 9, 12) &"(t) = (12, o, o 112 (till = √ (6+2) 2 + (12)2 + (12+)2 = √36+4 + 144 tunt? 11 al till = 360 * 4 + 4 + 4 +2) 112 (till = √6²(x+2) = 6.6*²42) to FCH (6+2, 12, 12t> 6 (4*t). < +2,2, 2t> (t?+2)

T'(4) = (4 42). 24 - £.(2 toon + (4²42) 0 - 2 12 t) } (t + 2)2 (+²+2) 2 (422). 2-24 ( 24 tuh + ² + 2)2 6 + ² +4 +- 2 ts) i + -ut) } + (2 +² +4-4+2)k (+²+2) 2 4t î . ut å + it2 +22 = 14+)2 + (-44)²+(4-2 + 2)2 6 + +4 +-22 12 +212 2 +2) T'CHI= + (9-2+²) Å 1 +212 +² +22 " T' (A) I1 (+²+2 of 1672 +164 + 16+ uth-16t (t² + 2) of ( ut2 + 4 + +4 ] (+²+2,4 4 & 5 = 2/ (+²+2)? U TG+) 11 2-**) -2t , 2.2t = (+2+2)2 et ty +²+2)2 N (A) U till TICA) 214272)2 <at -2t, 2-24 (t² + 2)2 (+242)2 (+2+2) 2 14--212 22 "ity 227, -2t, 2-2 > ? N(t) = lat, -2t

42 43 î tî hat + 2+2) th (-22²-4t) B (t) = (2x + 4 214) X" = 6+ 12 126 - î [144-0] } [72¢² - 144x2] th Co-1444] = nuuê +72+2 } -luuth 11 d' (t) x 2"(t) 11 = (144)2 + (72 + 2)2 + 144 +) = 12" +12?x6** *4 + 12°/2 = 12 12x12 + 36*4 + 12x122 = 12 62x4 + 62th & 6² x 442 = 12x6 Sut tht u t2 - 72 ist +22 = 72(+2+2) 1/2'G) xa" 72(+²+2) - Ex(425) (6(+242)] 36x6 (+ 2+2 72 HC) 3 3 (²+2) ²

kit)= 3 (+²+2) 2 ora or kate |T164) = 2/470" b 6(+²+2) 3(+²42)3 112 ll TE 2' (t). (2" (+) x 2" (A)) 12 (4) x 2" (A)/2 d . 12 2" (H) x 2" (t) = | 127 î Co-144] the Cor) = î [oro] - 144 &"(t) x 2" (4) = 2'(). («"(t) {d}"! (H) = (6+î +12 § +12th).( 1449) - 12x144 =17 28 2 (4) &&" (t) = 1 6+² 12 12t - î Co-o] - Î (o- 1997) fa (0-144] = runt } -1uuh 12 Lt) x 2 412 = (144+)2 + (-144 L = (144)² (at² +1) 1720 Lt L4GX1TG (+2+) 12 (+2+1) T: 12